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What's the locus of: $$\mathrm{Im} \bigg(\frac{z-z_{1}}{z-z_{2}}\bigg)=0$$

I tried to figure it out and I get that it's a line. But it looks to me it will be wrong.

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3 Answers 3

up vote 5 down vote accepted

Recall that $\frac{z-z_1}{z-z_2}$ is a complex number whose argument is the angle $z_1 z z_2$. Saying this number is real, is saying the angle is either $0$ or $\pi$, which means $z$ is on the line that contains $z_1$ and $z_2$.

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Put $z=x+iy,z_1=a+ib,z_2=c+id$

$$\frac{z-z_1}{z-z_2}=\frac{x-a+i(y-b)}{x-c+i(y-d)}=\frac{(x-a+i(y-b))(x-c-i(y-d))}{(x- c)^2+(y-d)^2}$$

$$Imz(\frac{z-z_1}{z-z_2})=\frac{(y-b)(x-c)-(y-d)(x-a)}{(x- c)^2+(y-d)^2}$$

$\implies (y-b)(x-c)=(y-d)(x-a)\implies x(d-b)+y(a-c)=ad-bc$

linear equation of $x,y\implies $ 2-D line.

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So it would be a line... –  Mykolas Sep 24 '12 at 19:10
    
If it would ask for real part being equal to 0, would it be a circle? @lab bhattacharjee –  Mykolas Sep 24 '12 at 19:23
    
@Mykolas, $x^2-(a+c)x+ac+y^2-(b+d)y-bd=0\implies $ 2-D circle. –  lab bhattacharjee Sep 24 '12 at 19:26
    
Thankyou very much @lab bhattacharjee –  Mykolas Sep 24 '12 at 19:26

Since $\Im\left(\frac{z-z_1}{z-z_2}\right)=0\Leftrightarrow \left(\frac{z-z_1}{z-z_2}\right)=c\in\mathbb{R}-\{0\}$, so for $z_1=a+b\cdot\mathbb{i}$ and $z_2=a'+b'\cdot\mathbb{i}$,

$z-z_1=c\cdot (z-z_2)\Leftrightarrow \left(x-a,y-b\right)=c\cdot\left(x-a',y-b'\right)\Leftrightarrow\begin{cases}x-a=c(x-a')\\y-b=c(y-b')\end{cases}$

and $\frac{x-a}{y-b}=\frac{x-a'}{y-b'}\Leftrightarrow y=-\frac{x-a}{a-a'}b'+b\frac{x-a'}{a-a'} $ which is a line.

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